//给定一个 m x n 二维字符网格 board 和一个字符串单词 word 。如果 word 存在于网格中，返回 true ；否则，返回 false 。
//
// 单词必须按照字母顺序，通过相邻的单元格内的字母构成，其中“相邻”单元格是那些水平相邻或垂直相邻的单元格。同一个单元格内的字母不允许被重复使用。
//
//
//
// 示例 1：
//
//
//输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word =
//"ABCCED"
//输出：true
//
//
// 示例 2：
//
//
//输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word =
//"SEE"
//输出：true
//
//
// 示例 3：
//
//
//输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word =
//"ABCB"
//输出：false
//
//
//
//
// 提示：
//
//
// m == board.length
// n = board[i].length
// 1 <= m, n <= 6
// 1 <= word.length <= 15
// board 和 word 仅由大小写英文字母组成
//
//
//
//
// 进阶：你可以使用搜索剪枝的技术来优化解决方案，使其在 board 更大的情况下可以更快解决问题？
// Related Topics 数组 回溯 矩阵 👍 1075 👎 0

package leetcode.editor.cn;

@SuppressWarnings("all")
//Java：单词搜索
public class 单词搜索 {
    public static void main(String[] args) {
        Solution solution = new 单词搜索().new Solution();
        // TO TEST

        char[][] board = {
                {'A', 'B', 'C', 'E'},
                {'S', 'F', 'C', 'S'},
                {'A', 'D', 'E', 'E'}};

        boolean exist = solution.exist(board, "ABCQ");
        System.out.println(exist);
    }

    //leetcode submit region begin(Prohibit modification and deletion)
    class Solution {
        public boolean exist(char[][] board, String word) {

            if (board == null || board.length == 0)
                return false;

            int line = board.length;
            int c = board[0].length;


            boolean exist = false;
            for (int i = 0; i < line; i++) {
                for (int j = 0; j < c; j++) {
                    if (word.charAt(0) == board[i][j]) {
                        exist = exist || dfs(board, word, i, j, 0);
                    }
                    if (exist)
                        return exist;
                }
            }
            return exist;
        }

        private boolean dfs(char[][] board, String word, int i, int j, int index) {
            int line = board.length;
            int c = board[0].length;


            if (board[i][j] != word.charAt(index)) {
                return false;
            }

            if (index == word.length() - 1) {
                return true;
            }

            if (i - 1 >= 0 && dfs(board, word, i - 1, j, index + 1))
                return true;
            if (i + 1 < line && dfs(board, word, i + 1, j, index + 1))
                return true;
            if (j - 1 >= 0 && dfs(board, word, i, j - 1, index + 1))
                return true;
            else return
                    j + 1 < c && dfs(board, word, i, j + 1, index + 1);

        }
    }
//leetcode submit region end(Prohibit modification and deletion)


}
